Billboard

Billboard

Time Limit : 20000/8000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 37   Accepted Submission(s) : 17

Problem Description

At the entrance to the university, there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is the place where all possible announcements are posted: nearest programming competitions, changes in the dining room menu, and other important information.

On September 1, the billboard was empty. One by one, the announcements started being put on the billboard.

Each announcement is a stripe of paper of unit height. More specifically, the i-th announcement is a rectangle of size 1 * wi.

When someone puts a new announcement on the billboard, she would always choose the topmost possible position for the announcement. Among all possible topmost positions she would always choose the leftmost one.

If there is no valid location for a new announcement, it is not put on the billboard (that's why some programming contests have no participants from this university).

Given the sizes of the billboard and the announcements, your task is to find the numbers of rows in which the announcements are placed.

 

Input

There are multiple cases (no more than 40 cases). The first line of the input file contains three integer numbers, h, w, and n (1 <= h,w <= 10^9; 1 <= n <= 200,000) - the dimensions of the billboard and the number of announcements. Each of the next n lines contains an integer number wi (1 <= wi <= 10^9) - the width of i-th announcement.

 

Output

For each announcement (in the order they are given in the input file) output one number - the number of the row in which this announcement is placed. Rows are numbered from 1 to h, starting with the top row. If an announcement can't be put on the billboard, output "-1" for this announcement.

 

Sample Input

3 5 5
2
4
3
3
3

 

Sample Output

1
2
1
3
-1

 

Author

hhanger@zju

 

Source

HDOJ 2009 Summer Exercise（5）

题意：

          有一个广告牌规格为h*w,h为高度，w为宽度，现在要往上面贴海报，海报的规格1*w,尽量往最上面的行贴同时尽量往左帖

           问每个海报的行数，贴不上去输出-1

 #include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define maxn 200005
int tree[maxn<<2];
int w;
int max(int a,int b)
{
    if(a<b) return b;
    return a;
}
void pushup(int rt)
{
    tree[rt]=max(tree[rt<<1],tree[rt<<1|1]);//向上更新每一行的最大容量
}
void build(int l,int r,int rt)
{
    tree[rt]=w;//初始化每一行的容量
    if(l==r) return ;
    int m=(l+r)/2;
    build(lson);
    build(rson);
}
int qurry(int x,int l,int r,int rt)//询问当前海报的位置
{
    if(l==r)//返回区间的边界即行数
        return l;
    int ret=0;
    int m=(l+r)>>1;
    if(x<=tree[rt<<1])//宽度比左儿子小，查询左二子
        ret=qurry(x,lson);
    else
        ret=qurry(x,rson);
    return ret;
}
void update(int x,int l,int r,int rt)
{
    if(l==r)//找到所在的行数
    {
        tree[rt]-=x;//对应节点的容量减去x
        return;
    }
    int m=(l+r)/2;
    if(x<=tree[rt<<1])//更新儿子
        update(x,lson);
    else update(x,rson);
    pushup(rt);//向上更新父节点
}
int main()
{
    int h,n,a;
    while(cin>>h>>w>>n)
    {
        if(h>n) h=n;//高度最大为n
        build(1,h,1);
        while(n--)
        {
            scanf("%d",&a);
            if(tree[1]<a)//头结点小于x
                cout<<-1<<endl;
            else
            {
                cout<<qurry(a,1,h,1)<<endl;
                update(a,1,h,1);
            }
        }
    }
}
//区间的边界表示的是行数，和以前不同的是输出的是区间的边界